Question

The British Department of Transportation studied to see if people avoid driving on Friday the 13th.  ...

The British Department of Transportation studied to see if people avoid driving on Friday the 13th.   They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations ("Friday the 13th," 2013). The data for each location on the two different dates is in following table:

Table: Traffic Count

Dates

6th

13th

1990, July

139246

138548

1990, July

134012

132908

1991, September

137055

136018

1991, September

133732

131843

1991, December

123552

121641

1991, December

121139

118723

1992, March

128293

125532

1992, March

124631

120249

1992, November

124609

122770

1992, November

117584

117263

Let ?1= mean traffic count on Friday the 6th. Let ?2 = mean traffic count on Friday the 13th. Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level.

(i) Determine the sample mean of differences x??di  Determine the sample mean of differences  x¯d

Enter in decimal form to nearest tenth.   

(ii) Determine sample standard deviation of differences sd

Enter in decimal form to nearest hundredth. Examples of correctly entered answers:  

0.01    0.20     0.35      3.00

(iii) Enter the level of significance ? used for this test:

Enter in decimal form. Examples of correctly entered answers:  0.01    0.02    0.05    0.10

(iv) Determine degrees of freedom for the sample of differences dfd:

Enter value in integer form. Examples of correctly entered answers:

2    5    9    23    77

(v) Determine t - score associated with critical value:  tc

Enter in decimal form to nearest thousandth. Examples of correctly entered answers:  

0.0011    0.020     0.500     0.371 2.000

(vi) Determine "error bound of the mean of difference" E

Enter value in decimal form rounded to nearest tenth. Examples of correctly entered answers:

2.0      0.3     1.6       11.7

(vii) Determine confidence interval of the mean difference ?d

Enter lower bound value to nearest tenth, followed by < , followed by "?d" for mean difference, followed by <, followed by upper bound value to nearest tenth. No spaces between any characters. Use "negative" sign if necessary. Do not use italics or enter units of measure. Examples of correctly entered answers:

0.74<?d<0.78

13.14<?d<13.96

-9.72<?d<-8.08

(viii) Using the confidence interval, select the correct description of the result of the survey:

A. We estimate with 90% confidence that the true population mean traffic count between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.

B. We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th falls outside 1154.1 and 2517.5.

C. We estimate with 90% confidence that the true sample mean traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.

D. We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.

Show Work.

Homework Answers

Answer #1

Solution

(i) Sample mean of differences: Dbar = 1835.8 ANSWER 1

(ii) Sample standard deviation of differences sD = 1176.01 ANSWER 2

(iii) Level of significance ? used for this test: 0.05 (5%) ANSWER 3

(iv) Degrees of freedom for the sample of differences dfd: = 9 ANSWER 4 [df = n - 1]

(v) t - score associated with critical value:  tc = 2.262 ANSWER 5

(vi) The "error bound of the mean of difference" E ± 841.3 ANSWER 6

(vii) 95% confidence interval of the mean difference ?d

        334.74 < ?d < 2017.28 ANSWER 7

DONE

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